B:(1,1),点B在抛物线上1=a*12a=1y=x2A:(2,0),B:(1,1)设直线AB的解析式为y=kx+b0=2k+b1=k+bk=-1 b=2y=-x+2y=x2x2+x-2=0(x-1)(x+2)=0x=1 y=1x=-2
y=4C:(-2,4)S△OBC=S△AOC-S△AOB
=1/2*2*4-1/2*2*1
=3S△AOD=3D:(X,x2)S△AOD=1/2*2*x2=3x=-根号3
x=根号3D:(-根号3,3)或者D:(根号3,3)两点间的距离=A:(x1,y1)B(x2,y2)AB=根号下[(x1-x2)^2+(y1-y2)^2]BC=根号下[(1-(-2))^2+(1-4)^2] =3倍根号2OC=根号下[(-2)^2+4^2] =2倍根号5BO=根号下[1^2+1^2] =根号2
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